OP Malhotra Class 11 Maths Solutions Chapter 18 Limits Exercise 18 (J)

Question 1. \( \text{Lim}_{x \to 0} \frac{(1+x)^m - 1}{(1+x)^n - 1} \)
Answer:The given limit is \( \text{Lt}_{x \to 0} \frac{(1+x)^m - 1}{(1+x)^n - 1} \). When we substitute \( x = 0 \), we get \( \frac{(1+0)^m - 1}{(1+0)^n - 1} = \frac{1 - 1}{1 - 1} = \frac{0}{0} \), which is an indeterminate form. So, we can use L'Hospital's Rule by taking the derivative of the numerator and the denominator separately with respect to \( x \).
\( \text{Lt}_{x \to 0} \frac{\frac{d}{dx}((1+x)^m - 1)}{\frac{d}{dx}((1+x)^n - 1)} \)
\( \implies \text{Lt}_{x \to 0} \frac{m(1+x)^{m-1} \cdot 1 - 0}{n(1+x)^{n-1} \cdot 1 - 0} \)
\( \implies \text{Lt}_{x \to 0} \frac{m(1+x)^{m-1}}{n(1+x)^{n-1}} \) Now, substitute \( x = 0 \) into the simplified expression:
\( \implies \frac{m(1+0)^{m-1}}{n(1+0)^{n-1}} \)
\( \implies \frac{m(1)^{m-1}}{n(1)^{n-1}} \)
\( \implies \frac{m \cdot 1}{n \cdot 1} = \frac{m}{n} \) So, the value of the limit is \( \frac{m}{n} \). This rule simplifies limits involving powers of \( (1+x) \).
In simple words: When you have a limit problem that looks like this and gives 0/0, you can solve it by finding the derivative of the top and bottom parts. After that, put the number for x back in to get your final answer.

🎯 Exam Tip: Always check if a limit results in an indeterminate form like \( \frac{0}{0} \) or \( \frac{\infty}{\infty} \) before applying L'Hospital's Rule; otherwise, the rule cannot be used.

Question 2. \( \text{Lim}_{x \to a} \frac{\log x - \log a}{x - a} \)
Answer:The given limit is \( \text{Lt}_{x \to a} \frac{\log x - \log a}{x - a} \). If we directly substitute \( x = a \), we get \( \frac{\log a - \log a}{a - a} = \frac{0}{0} \), which is an indeterminate form. Therefore, we can apply L'Hospital's Rule. We need to find the derivative of the numerator and the denominator with respect to \( x \).
\( \text{Lt}_{x \to a} \frac{\frac{d}{dx}(\log x - \log a)}{\frac{d}{dx}(x - a)} \)
\( \implies \text{Lt}_{x \to a} \frac{\frac{1}{x} - 0}{1 - 0} \)
\( \implies \text{Lt}_{x \to a} \frac{1}{x} \) Now, substitute \( x = a \) into the simplified expression:
\( \implies \frac{1}{a} \) So, the value of the limit is \( \frac{1}{a} \). This shows how L'Hospital's rule helps evaluate limits of logarithmic functions that lead to indeterminate forms.
In simple words: When you see this type of limit and get 0/0, you can find the derivative of the top part and the derivative of the bottom part. Then, put the 'a' back into 'x' to get the answer.

🎯 Exam Tip: Remember the derivatives of common functions, especially \( \log x \), when applying L'Hospital's Rule. Also, ensure the constant term's derivative is zero.

Question 3. \( \text{Lim}_{x \to a} \frac{x^a - a^x}{a^x - a^a} \)
Answer:The given limit is \( \text{Lt}_{x \to a} \frac{x^a - a^x}{a^x - a^a} \). When we substitute \( x = a \) directly, we get \( \frac{a^a - a^a}{a^a - a^a} = \frac{0}{0} \), which is an indeterminate form. So, we will use L'Hospital's Rule. This means we take the derivative of the numerator and the denominator with respect to \( x \).
\( \text{Lt}_{x \to a} \frac{\frac{d}{dx}(x^a - a^x)}{\frac{d}{dx}(a^x - a^a)} \) The derivative of \( x^a \) is \( a x^{a-1} \), and the derivative of \( a^x \) is \( a^x \log a \). The derivative of \( a^a \) (which is a constant) is 0.
\( \implies \text{Lt}_{x \to a} \frac{a x^{a-1} - a^x \log a}{a^x \log a} \) Now, substitute \( x = a \) into the expression:
\( \implies \frac{a \cdot a^{a-1} - a^a \log a}{a^a \log a} \)
\( \implies \frac{a^a - a^a \log a}{a^a \log a} \) We can factor out \( a^a \) from the numerator:
\( \implies \frac{a^a (1 - \log a)}{a^a \log a} \)
\( \implies \frac{1 - \log a}{\log a} \) This type of problem helps understand how L'Hospital's rule handles limits with mixed power and exponential terms.
In simple words: First, check if putting 'a' in place of 'x' gives 0/0. If it does, find the derivative of the top and bottom parts. Then, put 'a' back into 'x' for the new expression to find the final answer.

🎯 Exam Tip: Be careful with the derivatives of \( x^a \) (power rule) and \( a^x \) (exponential rule), as they are different and crucial for these types of limits.

Question 4. \( \text{Lim}_{x \to \frac{\pi}{6}} \frac{\sin(x - \frac{\pi}{6})}{\frac{\sqrt{3}}{2} - \cos x} \)
Answer:The given limit is \( \text{Lt}_{x \to \frac{\pi}{6}} \frac{\sin(x - \frac{\pi}{6})}{\frac{\sqrt{3}}{2} - \cos x} \). If we substitute \( x = \frac{\pi}{6} \), we get: Numerator: \( \sin(\frac{\pi}{6} - \frac{\pi}{6}) = \sin(0) = 0 \) Denominator: \( \frac{\sqrt{3}}{2} - \cos(\frac{\pi}{6}) = \frac{\sqrt{3}}{2} - \frac{\sqrt{3}}{2} = 0 \) Since we have the \( \frac{0}{0} \) indeterminate form, we can apply L'Hospital's Rule. We need to differentiate the numerator and denominator with respect to \( x \).
\( \text{Lt}_{x \to \frac{\pi}{6}} \frac{\frac{d}{dx}(\sin(x - \frac{\pi}{6}))}{\frac{d}{dx}(\frac{\sqrt{3}}{2} - \cos x)} \) The derivative of \( \sin(u) \) is \( \cos(u) \cdot u' \), and the derivative of \( -\cos x \) is \( -(-\sin x) = \sin x \).
\( \implies \text{Lt}_{x \to \frac{\pi}{6}} \frac{\cos(x - \frac{\pi}{6}) \cdot 1}{0 + \sin x} \)
\( \implies \text{Lt}_{x \to \frac{\pi}{6}} \frac{\cos(x - \frac{\pi}{6})}{\sin x} \) Now, substitute \( x = \frac{\pi}{6} \) into the simplified expression:
\( \implies \frac{\cos(\frac{\pi}{6} - \frac{\pi}{6})}{\sin(\frac{\pi}{6})} \)
\( \implies \frac{\cos(0)}{\sin(\frac{\pi}{6})} \) Since \( \cos(0) = 1 \) and \( \sin(\frac{\pi}{6}) = \frac{1}{2} \):
\( \implies \frac{1}{\frac{1}{2}} = 2 \) So, the value of the limit is \( 2 \). Understanding derivatives of trigonometric functions is key here.
In simple words: When you plug in the value for x and get 0/0, use L'Hospital's Rule. This means you find the derivative of the top and bottom parts. After that, put the x value back into the new expression to get the final answer.

🎯 Exam Tip: Always remember the standard values of trigonometric functions for common angles like \( \frac{\pi}{6} \) and the chain rule for derivatives involving functions of \( (x-a) \).

Question 5. \( \text{Lim}_{x \to a} \frac{\tan x - \tan a}{\sin a - \sin x} \)
Answer:The given limit is \( \text{Lt}_{x \to a} \frac{\tan x - \tan a}{\sin a - \sin x} \). If we substitute \( x = a \) directly, we get: Numerator: \( \tan a - \tan a = 0 \) Denominator: \( \sin a - \sin a = 0 \) This is the \( \frac{0}{0} \) indeterminate form, so we can use L'Hospital's Rule. We will differentiate the numerator and the denominator with respect to \( x \).
\( \text{Lt}_{x \to a} \frac{\frac{d}{dx}(\tan x - \tan a)}{\frac{d}{dx}(\sin a - \sin x)} \) The derivative of \( \tan x \) is \( \sec^2 x \), and \( \tan a \) is a constant, so its derivative is 0. The derivative of \( \sin a \) is 0, and the derivative of \( -\sin x \) is \( -\cos x \).
\( \implies \text{Lt}_{x \to a} \frac{\sec^2 x - 0}{0 - \cos x} \)
\( \implies \text{Lt}_{x \to a} \frac{\sec^2 x}{-\cos x} \) Now, substitute \( x = a \) into the simplified expression:
\( \implies \frac{\sec^2 a}{-\cos a} \) Since \( \sec a = \frac{1}{\cos a} \), we can rewrite the expression:
\( \implies \frac{1/\cos^2 a}{-\cos a} \)
\( \implies -\frac{1}{\cos^3 a} \)
\( \implies -\sec^3 a \) So, the value of the limit is \( -\sec^3 a \). This problem tests knowledge of derivatives of trigonometric functions and their reciprocal identities.
In simple words: When the limit gives 0/0, you need to use L'Hospital's Rule by taking the derivative of the top and bottom parts. Then, substitute 'a' for 'x' to find your answer.

🎯 Exam Tip: Be mindful of signs during differentiation (e.g., derivative of \( -\sin x \) is \( -\cos x \)) and simplify trigonometric expressions using identities like \( \sec x = 1/\cos x \).

Question 6. \( \text{Lim}_{x \to 4} \frac{3 - \sqrt{5+x}}{x - 4} \)
Answer:The given limit is \( \text{Lt}_{x \to 4} \frac{3 - \sqrt{5+x}}{x - 4} \). If we substitute \( x = 4 \): Numerator: \( 3 - \sqrt{5+4} = 3 - \sqrt{9} = 3 - 3 = 0 \) Denominator: \( 4 - 4 = 0 \) This is the \( \frac{0}{0} \) indeterminate form, so L'Hospital's Rule can be applied. We need to find the derivative of the numerator and the denominator with respect to \( x \).
\( \text{Lt}_{x \to 4} \frac{\frac{d}{dx}(3 - \sqrt{5+x})}{\frac{d}{dx}(x - 4)} \) The derivative of \( 3 \) is 0. For \( \sqrt{5+x} = (5+x)^{1/2} \), its derivative is \( \frac{1}{2}(5+x)^{-1/2} \cdot 1 \). The derivative of \( x-4 \) is \( 1-0 = 1 \).
\( \implies \text{Lt}_{x \to 4} \frac{0 - \frac{1}{2}(5+x)^{-\frac{1}{2}} \cdot (0+1)}{1 - 0} \)
\( \implies \text{Lt}_{x \to 4} -\frac{1}{2}(5+x)^{-\frac{1}{2}} \) Now, substitute \( x = 4 \) into the simplified expression:
\( \implies -\frac{1}{2}(5+4)^{-\frac{1}{2}} \)
\( \implies -\frac{1}{2}(9)^{-\frac{1}{2}} \) We know that \( (9)^{-\frac{1}{2}} = \frac{1}{\sqrt{9}} = \frac{1}{3} \).
\( \implies -\frac{1}{2} \cdot \frac{1}{3} = -\frac{1}{6} \) So, the value of the limit is \( -\frac{1}{6} \). This problem illustrates how to handle square roots in limits using derivatives effectively.
In simple words: When a limit gives 0/0, apply L'Hospital's Rule by taking the derivative of the top part and the bottom part. Then, substitute the value of x into the new expression to get the final answer.

🎯 Exam Tip: Remember that \( \sqrt{f(x)} = (f(x))^{1/2} \) and its derivative is \( \frac{1}{2}(f(x))^{-1/2}f'(x) \). Pay attention to negative signs and fractional exponents.

Question 7. \( \text{Lim}_{x \to 0} \frac{a^x - 1}{b^x - 1} \)
Answer:The given limit is \( \text{Lt}_{x \to 0} \frac{a^x - 1}{b^x - 1} \). If we substitute \( x = 0 \) directly: Numerator: \( a^0 - 1 = 1 - 1 = 0 \) Denominator: \( b^0 - 1 = 1 - 1 = 0 \) This is the \( \frac{0}{0} \) indeterminate form, so L'Hospital's Rule is applicable. We will differentiate the numerator and the denominator with respect to \( x \).
\( \text{Lt}_{x \to 0} \frac{\frac{d}{dx}(a^x - 1)}{\frac{d}{dx}(b^x - 1)} \) The derivative of \( a^x \) is \( a^x \log a \), and the derivative of \( b^x \) is \( b^x \log b \). The derivative of constants (1) is 0.
\( \implies \text{Lt}_{x \to 0} \frac{a^x \log a - 0}{b^x \log b - 0} \)
\( \implies \text{Lt}_{x \to 0} \frac{a^x \log a}{b^x \log b} \) Now, substitute \( x = 0 \) into the simplified expression:
\( \implies \frac{a^0 \log a}{b^0 \log b} \) Since \( a^0 = 1 \) and \( b^0 = 1 \):
\( \implies \frac{1 \cdot \log a}{1 \cdot \log b} \)
\( \implies \frac{\log a}{\log b} \) So, the value of the limit is \( \frac{\log a}{\log b} \). This result is a fundamental limit identity often seen in calculus.
In simple words: When you have a limit that gives 0/0, use L'Hospital's Rule. This means finding the derivative of the top and bottom parts. Then, put the value of x back in to get your final answer.

🎯 Exam Tip: Memorize the derivative formula for \( a^x \), which is \( a^x \log a \), as it's a common step in exponential limit problems.

Question 8. \( \text{Lim}_{\theta \to \frac{\pi}{2}} \frac{1 - \sin \theta}{\cos \theta} \)
Answer:The given limit is \( \text{Lt}_{\theta \to \frac{\pi}{2}} \frac{1 - \sin \theta}{\cos \theta} \). If we substitute \( \theta = \frac{\pi}{2} \) directly: Numerator: \( 1 - \sin(\frac{\pi}{2}) = 1 - 1 = 0 \) Denominator: \( \cos(\frac{\pi}{2}) = 0 \) This is the \( \frac{0}{0} \) indeterminate form, so L'Hospital's Rule can be applied. We need to find the derivative of the numerator and the denominator with respect to \( \theta \).
\( \text{Lt}_{\theta \to \frac{\pi}{2}} \frac{\frac{d}{d\theta}(1 - \sin \theta)}{\frac{d}{d\theta}(\cos \theta)} \) The derivative of \( 1 \) is 0. The derivative of \( -\sin \theta \) is \( -\cos \theta \). The derivative of \( \cos \theta \) is \( -\sin \theta \).
\( \implies \text{Lt}_{\theta \to \frac{\pi}{2}} \frac{-\cos \theta}{-\sin \theta} \) We can simplify the expression:
\( \implies \text{Lt}_{\theta \to \frac{\pi}{2}} \frac{\cos \theta}{\sin \theta} \)
\( \implies \text{Lt}_{\theta \to \frac{\pi}{2}} \cot \theta \) Now, substitute \( \theta = \frac{\pi}{2} \) into the simplified expression:
\( \implies \cot(\frac{\pi}{2}) \) Since \( \cot(\frac{\pi}{2}) = 0 \):
\( \implies 0 \) So, the value of the limit is \( 0 \). This demonstrates how to simplify the expression before final evaluation.
In simple words: When a limit gives 0/0, use L'Hospital's Rule to take the derivative of the top and bottom parts. After simplifying the expression, substitute the value for theta to get the final answer.

🎯 Exam Tip: Simplify the trigonometric expression (like \( \cos \theta / \sin \theta = \cot \theta \)) after differentiation and before substituting the limit value to avoid errors.

Question 9. \( \text{Lim}_{x \to \infty} \frac{x}{2^x} \)
Answer:The given limit is \( \text{Lt}_{x \to \infty} \frac{x}{2^x} \). If we substitute \( x = \infty \) directly: Numerator: \( \infty \) Denominator: \( 2^\infty = \infty \) This is the \( \frac{\infty}{\infty} \) indeterminate form, so L'Hospital's Rule can be applied. We need to find the derivative of the numerator and the denominator with respect to \( x \).
\( \text{Lt}_{x \to \infty} \frac{\frac{d}{dx}(x)}{\frac{d}{dx}(2^x)} \) The derivative of \( x \) is \( 1 \). The derivative of \( 2^x \) is \( 2^x \log 2 \).
\( \implies \text{Lt}_{x \to \infty} \frac{1}{2^x \log 2} \) Now, substitute \( x = \infty \) into the simplified expression:
\( \implies \frac{1}{2^\infty \log 2} \) Since \( 2^\infty = \infty \) and \( \log 2 \) is a positive constant:
\( \implies \frac{1}{\infty \cdot \log 2} \)
\( \implies \frac{1}{\infty} = 0 \) So, the value of the limit is \( 0 \). This result shows that exponential functions grow much faster than linear functions, making the fraction approach zero.
In simple words: When you have a limit that gives infinity over infinity, use L'Hospital's Rule to take the derivative of the top and bottom parts. Then, plug in infinity for x to find your answer.

🎯 Exam Tip: Remember that exponential functions like \( a^x \) grow faster than polynomial functions like \( x \), so their limits at infinity often result in 0 or infinity depending on their position in the fraction.

Question 10. \( \text{Lim}_{x \to \frac{\pi}{2}} \frac{1 - \sin x}{x \cos x} \)
Answer:The given limit is \( \text{Lt}_{x \to \frac{\pi}{2}} \frac{1 - \sin x}{x \cos x} \). If we substitute \( x = \frac{\pi}{2} \) directly: Numerator: \( 1 - \sin(\frac{\pi}{2}) = 1 - 1 = 0 \) Denominator: \( \frac{\pi}{2} \cos(\frac{\pi}{2}) = \frac{\pi}{2} \cdot 0 = 0 \) This is the \( \frac{0}{0} \) indeterminate form, so L'Hospital's Rule can be applied. We need to find the derivative of the numerator and the denominator with respect to \( x \).
\( \text{Lt}_{x \to \frac{\pi}{2}} \frac{\frac{d}{dx}(1 - \sin x)}{\frac{d}{dx}(x \cos x)} \) The derivative of \( 1 - \sin x \) is \( 0 - \cos x = -\cos x \). For the denominator, we use the product rule: \( \frac{d}{dx}(uv) = u'v + uv' \). Here, \( u=x \) and \( v=\cos x \). So, \( \frac{d}{dx}(x \cos x) = (1)(\cos x) + (x)(-\sin x) = \cos x - x \sin x \).
\( \implies \text{Lt}_{x \to \frac{\pi}{2}} \frac{-\cos x}{\cos x - x \sin x} \) Now, substitute \( x = \frac{\pi}{2} \) into the simplified expression:
\( \implies \frac{-\cos(\frac{\pi}{2})}{\cos(\frac{\pi}{2}) - \frac{\pi}{2} \sin(\frac{\pi}{2})} \) Since \( \cos(\frac{\pi}{2}) = 0 \) and \( \sin(\frac{\pi}{2}) = 1 \):
\( \implies \frac{-0}{0 - \frac{\pi}{2} \cdot 1} \)
\( \implies \frac{0}{-\frac{\pi}{2}} = 0 \) So, the value of the limit is \( 0 \). This problem highlights the importance of using the product rule correctly when differentiating the denominator.
In simple words: When the limit gives 0/0, use L'Hospital's Rule. This means taking the derivative of the top and bottom parts separately. For the bottom, use the product rule. Then, put the value of x back in to get your final answer.

🎯 Exam Tip: Remember to apply the product rule for differentiation when the denominator or numerator is a product of functions (e.g., \( x \cos x \)) before substituting the limit value.